AVR-Chat

Yes, to be accurate, the solenoid would be connected to "+12V IN". 
I'm given one terminal to ground, the other is already connected to
+12V IN.  Connecting the diode on the ground line would provide me
with the opposite problem, negative spikes (i.e. from ESD, or what
have you) would not be taken care of.  I'll take a look.

--- In AVR-Chat@yahoogroups.com, "Steve Baldwin" <steve@...> wrote:
>
> Err. OK. By "Exactly correct" you mean that it isn't as you've shown
it in your 
> updated diagram at http://i5.tinypic.com/20hrl1i.jpg. Correct ?
> 
> You said that the wiring between the solenoid and your circuit is
integrated in 
> some way. I can't remember the details. If you provide 2 screw
terminals (or 
> similar) for the user to connect the solenoid and the connection to
the supply 
> is within your circuit/pcb/box, I would make that positive terminal
"+12V", 
> rather than "+12V In". ie. As you've drawn it.
> You get D2 where you want it and the solution is conventional. All
is good.
> 
> If however, you provide a single terminal and the user wires up the
positive 
> end of the solenoid, I would take D3 and put it facing the other
direction in 
> the negative supply. You get D2 across the load still and reverse
polarity 
> protection. The downside is if you have any external inputs that are
ground 
> referenced to this circuit, there will be a voltage drop across the
diode. The 
> other downside is that it is non-conventional. That means that
someone else 
> looking at it would scratch their head for a while. IMHO, you should
avoid 
> head-scratchers unless there is good reason to use one.
> 
> If that is the case, then you'd need to look at using a FET as a
zero drop 
> diode. Google should have something.
> 
> Steve.
> 
> 
> 
> On 20 Jul 2006 at 15:57, Ryan wrote:
> 
> > Exactly correct - the inductor is connected to +12V In.  D3 is for
> > reverse polarity protection to the driver circuit.  D2 is supposed to
> > kill any voltage spikes, but would be blocked by D3 as shown.  Therein
> > lies the problem.
> > 
> > Hopefully this gets through ok, yahoo has been having problems with
> > their site.
> > 
> > 
> > --- In AVR-Chat@yahoogroups.com, "Steve Baldwin" <steve@> wrote: >
> > > Where's the problem ? > As you've drawn it, D2 is across the
> > inductor. Just where it should be. > You said that the wiring to the
> > inductor is integral to your unit so the only > place you can get the
> > polarity reversed is at the "+12V In" node. There's no > hinderance
> > there unless you have drawn the circuit wrong and the inductor > is
> > connected to  "+12V In". > > Steve. > > >
> > ========================================== > Steve Baldwin            
> >              Electronic Product Design > TLA Microsystems Ltd         
> >    Microcontroller Specialists > PO Box 15-680, New Lynn              
> >  http://www.tla.co.nz > Auckland, New Zealand                     ph 
> > +64 9 820-2221 > email: steve@                      fax +64 9
> > 820-1929 > ========================================= >
> 
> ==========================================
> Steve Baldwin                          Electronic Product Design
> TLA Microsystems Ltd             Microcontroller Specialists
> PO Box 15-680, New Lynn                http://www.tla.co.nz
> Auckland, New Zealand                     ph  +64 9 820-2221
> email: steve@...                      fax +64 9 820-1929
> =========================================
>